Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
//package here;

/**
 *
 * @author sokumar
 */
import java.util.*;
import java.io.*;
public class Main {
   
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter w = new PrintWriter(System.out);
        
         int n = sc.nextInt();
         double []p = new double[n];
         for(int i=0;i<n;i++)
         {
             p[i] = sc.nextDouble();
         }
         double [][]dp = new double[n+1][n+1];
         dp[1][1] = p[0];
         dp[1][0] = 1.0-p[0];
         for(int i=2;i<=n;i++)
         {
             for(int j=0;j<=i;j++)
             {
                 if(j!=0)
                 dp[i][j] = dp[i-1][j]*(1-p[i-1]) + dp[i-1][j-1]*p[i-1];
                 else
                     dp[i][j] = dp[i-1][j]*(1-p[i-1]);
                 //w.println(i + " "+ j + " " + dp[i][j]);
             }
         }
         double res = 0;
       
             for(int j=n/2 + 1;j<=n;j++)
             {
                 res += dp[n][j];
                 
             }
         
         w.println(res);
         
            
        w.close();
    }
    
    
}